In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets
AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH = CK.

Question

04-10-2017
Mathematics

contestada

In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets
AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH = CK.

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meerkat18 meerkat18 · Answer 1 · 2017-10-17T19:08:16+02:00

First, let's illustrate the problem as shown in the picture attached. Then, let's make a two-column proof:

Statement Reason
m∠B = m∠C Given
AB = AC Isosceles Triangle Definition
AK=KB=AH=HC Angle Bisector Postulate
ΔBKC = ΔHBC Line BC is common side
BH = CK Corresponding parts of congruent triangles are congruent

In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH = CK.

Respuesta :

Otras preguntas

In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets
AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH = CK.