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Altitudes and the Orthocenter: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two angle parts made by an altitude

Triangle ABC is acute. One base angle is $\angle ABC=58^\circ$ .

The other base angle is $\angle ACB=47^\circ$ .

The altitude from A meets BC at D. Find angles BAD and DAC. Decide whether AD is also the internal angle bisector from A, and justify your decision.

AD is perpendicular to BC, with D between B and C.

Text description of this figure

The given base angles of triangle ABC are 58 degrees and 47 degrees. AD is perpendicular to BC at an interior point D. Neither requested vertex-angle part is labeled.
1. Hint 1
  The altitude makes two right triangles sharing AD.
2. Hint 2
  Find the remaining acute angle in each right triangle, then compare the two angle parts.
Answer

$\begin{aligned}\angle BAD&=32^\circ\\\angle DAC&=43^\circ.\end{aligned}$
AD is not the internal angle bisector.

Full solution

Since AD is perpendicular to BC, triangle ABD is right at D.
$\begin{aligned}\angle BAD&=90^\circ-58^\circ\\\angle BAD&=32^\circ.\end{aligned}$

Triangle ACD is also right at D.
$\begin{aligned}\angle DAC&=90^\circ-47^\circ\\\angle DAC&=43^\circ.\end{aligned}$

The two angle parts differ, so AD does not bisect angle BAC.
Perpendicularity to BC alone does not imply equal vertex-angle parts.

Answer

$\begin{aligned}\angle BAD&=32^\circ\\\angle DAC&=43^\circ.\end{aligned}$
AD is not the internal angle bisector.

Key idea

An altitude supplies right angles at its foot, not automatically equal angles at its vertex.
Problem 2 Construct all three lines in an obtuse case

Draw a scalene triangle ABC with its angle at A obtuse. Using a compass and unmarked straightedge, construct all three altitude lines and label their common point H.

Keep the construction arcs. Label the feet D, E, F from A, B, C respectively, mark the right angles, and show every necessary side or altitude extension. Explain why the constructed lines have the required perpendicular conditions.
1. Hint 1
  Use the full supporting side line when constructing a perpendicular from a vertex.
2. Hint 2
  A circle centered at the vertex with a radius large enough to cut that line twice supplies two points equally far from the vertex.
Answer

A valid construction has AD perpendicular to BC, BE perpendicular to CA, and CF perpendicular to AB. E and F lie beyond A on the side extensions; the full altitude lines meet at exterior point H.

Full solution

For the altitude from B, extend AC past A.
Draw a circle centered at B with radius greater than BA, meeting line AC at distinct U and V.
Construct the perpendicular bisector of UV; it contains B because BU and BV are equal radii.
Its intersection with line AC is the foot E.

Repeat the perpendicular-from-a-point construction from A to line BC and from C to line AB.
Label the feet D and F and retain the arcs that establish the perpendiculars.

The altitude-concurrence theorem makes the three complete lines meet at H.
In this obtuse case, E and F lie beyond A.
Extending the altitude lines locates H outside the triangle.

The overview shows the final lines; keep your compass arcs on your own drawing.

Answer

A valid construction has AD perpendicular to BC, BE perpendicular to CA, and CF perpendicular to AB. E and F lie beyond A on the side extensions; the full altitude lines meet at exterior point H.

Key idea

Construct on complete side lines first, then distinguish feet from the common intersection.
Problem 3 Establish an auxiliary midpoint

Start with nondegenerate triangle ABC. Through A draw the parallel to BC, through B the parallel to CA, and through C the parallel to AB. Their intersections form PQR, with A between Q and R, B between R and P, and C between P and Q.

Prove that B is the midpoint of RP by identifying two parallelograms and the opposite-side equalities they supply. Do not assume the midpoint conclusion from the drawing.

Each auxiliary side passes through one original vertex.

Text description of this figure

Through A, B, C, parallels to BC, CA, AB form auxiliary triangle PQR. A is on QR, B on RP, and C on PQ. No equal halves or perpendicular conclusions are marked.
1. Hint 1
  Find one parallelogram containing RB and another containing BP.
2. Hint 2
  The two required lengths can each be compared with AC.
Answer

B is the midpoint of RP; parallelograms ACBR and ABPC give RB and BP the same length as AC.

Full solution

In ACBR, AC is parallel to BR and CB is parallel to RA, by the construction.
Thus ACBR is a parallelogram, and its opposite sides give
$RB=AC.$

In ABPC, AB is parallel to PC and BP is parallel to CA.
Thus ABPC is a parallelogram, and its opposite sides give
$BP=AC.$

Consequently $RB=BP$ .
The supplied order puts B between R and P.
Equal halves on that segment make B its midpoint.

The double-tick pair records the midpoint established in this problem.

Answer

B is the midpoint of RP; parallelograms ACBR and ABPC give RB and BP the same length as AC.

Key idea

A midpoint conclusion needs both equal lengths and the point between the endpoints.
Problem 4 One interior foot does not settle the location

In triangle ABC, the altitude from A has its foot D strictly inside side BC. Does this alone force the orthocenter to lie inside the triangle?

Compare an acute triangle with a triangle obtuse at A. Explain what happens to D and to the orthocenter in each case.

The supplied fact concerns D, not the common point H.

Text description of this figure

The given drawing has one perpendicular AD with its foot D inside BC. It does not mark an orthocenter or claim a general location rule.
1. Hint 1
  The foot of the altitude from an obtuse vertex behaves differently from the other two feet.
2. Hint 2
  Compare the full angle type of the triangle with the information about just one foot.
Answer

No. An acute triangle has both D and H inside the indicated side/triangle. A triangle obtuse at A still has D inside BC but has H outside the triangle.

Full solution

In an acute triangle, all three feet lie inside the opposite sides, and two altitude segments cross inside.
Thus D is inside BC and H is inside the triangle.

If A is obtuse, the angles at B and C are acute.
Their acute directions still put the perpendicular foot D from A between B and C.
However, the feet from B and C lie beyond A on the side extensions.

For example, the line from B through its exterior foot on AC meets the triangle only at B.
Its crossing with the altitude from A is not B, so H is outside.
The information about D alone therefore allows both an interior and an exterior orthocenter.

This obtuse case satisfies the interior-foot condition but has an exterior H.

Answer

No. An acute triangle has both D and H inside the indicated side/triangle. A triangle obtuse at A still has D inside BC but has H outside the triangle.

Key idea

A single altitude foot does not determine the orthocenter location; the triangle angle type does.
Problem 5 When one segment has two supplied roles

In nondegenerate triangle ABC, D lies strictly inside BC. AD is an altitude and also the internal bisector of angle BAC.

Prove that AD is a median. State the triangle-congruence test and the corresponding sides that establish your conclusion.

AD is perpendicular to BC and bisects angle BAC.

Text description of this figure

AD is perpendicular to BC at D. Matching single arcs show that AD bisects the angle at A. No equal side halves are marked before the proof.
1. Hint 1
  Compare the two right triangles on either side of AD.
2. Hint 2
  They have equal angles at A, equal right angles at D, and a common side between those angles.
Answer

AD is a median; ASA gives congruent triangles BAD and CAD, and the corresponding sides BD and DC are equal.

Full solution

The altitude makes angles BDA and CDA right.
The internal-bisector condition gives equal angles BAD and CAD.
The included side AD is common.

ASA therefore gives congruent triangles BAD and CAD.
The two nondegenerate triangles have the correspondence B to C, A to A, and D to D.

Corresponding sides give $BD=DC$ .
Since D lies between B and C, it is the midpoint of BC.
Thus AD joins A to the opposite midpoint and is a median.

The side-half equality is a conclusion, so it appears only with the solution.

Answer

AD is a median; ASA gives congruent triangles BAD and CAD, and the corresponding sides BD and DC are equal.

Key idea

Two supplied roles can force a third, but the equality needs a congruence argument.
Problem 6 Annotate the auxiliary correspondence

The given figure has A, B, C as the midpoints of auxiliary sides QR, RP, PQ. The parallel pairs are QR with BC, RP with CA, and PQ with AB.

Copy the figure on paper. Draw the three original altitude lines, mark their right angles with the auxiliary sides, and label the common point H.

For each line, identify which original altitude and which auxiliary perpendicular bisector it represents. Explain how the parallels justify your right-angle marks.

The supplied midpoint facts belong to the auxiliary triangle.

Text description of this figure

A, B, C are the established midpoints of auxiliary sides QR, RP, PQ. Those sides are parallel to BC, CA, AB respectively. No altitude lines or common center are marked.
1. Hint 1
  A line perpendicular to an original side is perpendicular to its parallel auxiliary side.
2. Hint 2
  For each auxiliary side, use the supplied midpoint as the vertex through which to draw the perpendicular.
Answer

The altitude from A is the perpendicular bisector of QR; the altitude from B is the perpendicular bisector of RP; the altitude from C is the perpendicular bisector of PQ. They meet at H.

Full solution

Draw the perpendicular through A to BC.
Since QR is parallel to BC, it also makes a right angle with QR at its midpoint A.
Label this line as the altitude from A and the perpendicular bisector of QR.

Draw the perpendicular through B to CA.
Since RP is parallel to CA, this line is the perpendicular bisector of RP.
Draw the perpendicular through C to AB; parallelism similarly makes it the perpendicular bisector of PQ.

The perpendicular-bisector concurrence theorem for PQR puts the three lines through a common point H.
This is the orthocenter of ABC.

Each line is a perpendicular bisector of an auxiliary side.

Answer

The altitude from A is the perpendicular bisector of QR; the altitude from B is the perpendicular bisector of RP; the altitude from C is the perpendicular bisector of PQ. They meet at H.

Key idea

The same drawn line has an altitude role in one triangle and a perpendicular-bisector role in another.
Problem 7 Find the first invalid line and repair the proof

In nondegenerate triangle ABC, AD is perpendicular to BC at D. Consider this proposed argument.

Line 1. AD is an altitude from A.

Line 2. D is the midpoint of BC because an altitude always bisects its opposite side.

Line 3. Line AD is therefore the perpendicular bisector of BC.

Line 4. Applying this to all three altitudes makes the orthocenter the circumcenter of ABC.

Identify the first invalid line and explain what is missing. Then describe how the auxiliary triangle repairs the concurrence proof without assuming midpoint feet on the original sides.

The figure records the supplied perpendicular condition.

Text description of this figure

Only the given right angle between AD and BC is marked. No equality ticks on BD and DC are provided.
1. Hint 1
  Compare the information supplied by a right-angle mark with the information supplied by a midpoint mark.
2. Hint 2
  In the correct proof, ask which triangle has the original vertices as its side midpoints.
Answer

Line 2 is first invalid. Altitudes need not bisect original sides. The repaired proof makes the altitude lines perpendicular bisectors of auxiliary sides, concurrent at the auxiliary circumcenter.

Full solution

Line 1 follows from the given perpendicular from A to its opposite side line.
Line 2 has no midpoint justification: perpendicularity alone does not give equal lengths BD and DC.
This is the first invalid step.

Construct the three parallels through the original vertices to the opposite sides.
Parallelogram properties show that the original vertices A, B, C are the midpoints of auxiliary sides QR, RP, PQ.

An original altitude line is perpendicular to the corresponding auxiliary side because that side is parallel to the original opposite side.
It also passes through the proved auxiliary midpoint.
The three lines are therefore the perpendicular bisectors of PQR and concur at its circumcenter, the orthocenter of ABC.
The argument does not generally make that point the original circumcenter.

Each line is a perpendicular bisector of an auxiliary side.

Answer

Line 2 is first invalid. Altitudes need not bisect original sides. The repaired proof makes the altitude lines perpendicular bisectors of auxiliary sides, concurrent at the auxiliary circumcenter.

Key idea

Perpendicularity and midpoint placement are separate hypotheses in a perpendicular-bisector proof.
Problem 8 Reverse the transfer through a medial triangle

In nondegenerate triangle ABC, D, E, F are the midpoints of BC, CA, AB respectively. Join them to form the medial triangle DEF.

Prove that the perpendicular bisectors of BC, CA, AB are exactly the altitude lines from D, E, F respectively in triangle DEF.

Identify the orthocenter of DEF in terms of the circumcenter of ABC. Justify that DEF is nondegenerate and that the common point exists and is unique. Allow supporting side-line extensions wherever needed.

D, E, F are the supplied side midpoints; no center is marked.

Text description of this figure

D, E, F are the midpoints of BC, CA, AB, marked by distinct one-, two-, and three-tick pairs. The midpoint segments form triangle DEF. No parallel marks, perpendicular lines, feet or centers are supplied.
1. Hint 1
  Compare each medial side with an original side.
2. Hint 2
  Locate an original perpendicular bisector using its midpoint, then compare its direction with the opposite medial side.
3. Hint 3
  After matching all three complete lines, apply a known concurrence theorem.
Answer

The perpendicular bisectors of BC, CA, AB are the altitude lines from D, E, F of DEF. Their unique common point is both the circumcenter of ABC and the orthocenter of DEF.

Full solution

The midsegment theorem gives
$\begin{aligned}EF&\parallel BC\\FD&\parallel CA\\DE&\parallel AB.\end{aligned}$

Each medial side has half the positive length of its corresponding original side, so D, E, F are distinct.
Lines EF and FD are parallel to the nonparallel lines BC and CA, so they are nonparallel.
Hence DEF is a nondegenerate triangle.

D is the midpoint of BC.
Its original perpendicular bisector is the unique line through D perpendicular to BC.
Since EF is parallel to BC, this same line is perpendicular to EF.
Thus it is exactly the altitude line from D in DEF; uniqueness of the perpendicular through D identifies the full lines.

E is the midpoint of CA, and FD is parallel to CA.
The unique perpendicular through E to CA is therefore exactly the altitude line from E to supporting line FD.

F is the midpoint of AB, and DE is parallel to AB.
The unique perpendicular through F to AB is therefore exactly the altitude line from F to supporting line DE.

Use the complete supporting lines EF, FD, DE to locate the altitude feet.
A foot can lie outside its medial side segment, in which case that side line must be extended.
The common point need not lie inside either triangle.

Because ABC is nondegenerate, its three perpendicular bisectors have exactly one common point, its circumcenter O.
These are exactly the three altitude lines of DEF, so their unique common point is also its orthocenter.
Thus the orthocenter of DEF is O, the circumcenter of ABC.

The same three complete lines identify the original circumcenter and medial orthocenter.

Answer

The perpendicular bisectors of BC, CA, AB are the altitude lines from D, E, F of DEF. Their unique common point is both the circumcenter of ABC and the orthocenter of DEF.

Key idea

Midsegments turn the original side perpendicular bisectors into the altitude lines of the medial triangle.
Problem 9 Separate the centers of a right triangle

Triangle ABC is right at C. Let O, I, G, and H denote its circumcenter, incenter, centroid, and orthocenter respectively.

Describe the location of each point. Explain why H differs from each of the other three centers, without calculating coordinates or lengths.

ABC is right at C; no centers are marked.

Text description of this figure

Triangle ABC is right at C. The statement supplies four center names, but none is marked in the givens figure.
1. Hint 1
  Use the two legs as altitude lines, then recall the circumcenter location in a right triangle.
2. Hint 2
  The internal-bisector and median results place their centers strictly inside a nondegenerate triangle.
Answer

H is vertex C. O is the midpoint of hypotenuse AB. I and G lie strictly inside the triangle. Thus H is distinct from O, I, and G.

Full solution

Because AC and BC are perpendicular, they are two altitude lines, meeting at C.
Hence H is C.

The right-triangle circumcenter theorem puts O at the midpoint of AB.
This point lies on the hypotenuse, whereas C is off the hypotenuse, so O differs from H.

The incenter I lies strictly inside by the internal-angle-bisector concurrence theorem.
The centroid G lies strictly inside by the median theorem.
Neither interior point is the vertex C, so H differs from both.

The right-angle vertex is separate from the other three centers.

Answer

H is vertex C. O is the midpoint of hypotenuse AB. I and G lie strictly inside the triangle. Thus H is distinct from O, I, and G.

Key idea

Each center follows its defining theorem; one special triangle need not make them coincide.
Problem 10 What a shared centroid and orthocenter force

In nondegenerate triangle ABC, its centroid G and orthocenter H are the same point. Prove that the triangle is equilateral.

Then explain why the circumcenter and incenter must also be this point. Give a geometric justification.
1. Hint 1
  A line from a vertex through the shared interior point is both a median line and an altitude line.
2. Hint 2
  Compare the two triangles on either side of a median that is also perpendicular to its opposite side.
3. Hint 3
  Repeat from a second vertex so that the two resulting side equalities connect all three sides.
Answer

ABC is equilateral, and its circumcenter, incenter, centroid, and orthocenter all coincide at the supplied point.

Full solution

Let D be the midpoint of BC.
The centroid lies inside median AD.
Because G is also H, the altitude line from A contains the same two distinct points A and G.
Thus it is line AD, and AD is perpendicular to BC.

Triangles ADB and ADC have the equal side halves BD and DC, the common side AD, and equal included right angles at D.
SAS gives their congruence, so AB and AC are equal.

Let E be the midpoint of AC.
The same argument makes median BE an altitude line.
SAS applied to BEA and BEC gives BA and BC equal.
Together the two equalities make all three sides equal.

In an equilateral triangle each median is also an altitude, internal angle bisector, and perpendicular bisector, as proved in the lesson.
Their common point is therefore also the circumcenter and incenter.

The extra center coincidences follow from the equilateral result.

Answer

ABC is equilateral, and its circumcenter, incenter, centroid, and orthocenter all coincide at the supplied point.

Key idea

Coinciding defining constructions can force a special triangle, rather than merely occur in one.

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