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:

for exempl this   system of equations   y = x - 3

                                                                y = 2x - 5

look   the line : y = x -3     ( red)

                            the line : y = 2x -5 ( bleus)