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.

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

step-by-step explanation:

my butt

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: