Question

if BD and CE bisect angle B and angle C in triangle ABC, where AB=AC. Show that triangle BDC = congruent triangle CEB​

Answer 1

GIVEN: in triangle ABC,

AB=AC

BD & EC are 2 median

TO PROOF:

triangle BCD cong. to triangle CEB

PROOF:

In triangle ABC,

AB=AC(given)

=>Therefore, triangle ABC is an isos.triangle

angle EBC = angle DCB ( base angle of isos. triangle are equal) ---[1(MARK AS)]

Also,1/2 AB=1/2 AC,

=>BE=CD---[2(MARK AS)]

So,

In triangle BCD&CEB

BC=BC(common)

angle EBC=angle DCB(from(2))

BF=CD(from (2))

Therefore,

Triangle BCD is cong.to CEB(by SAS cong. rule)

Hence Proved

Answer 2

Proof :

AB=AC {GIVEN]

angle B=angle C ( GIVEN]

In triangle BD=BC (COMMON]

angle C= angle B

angle 1 =angle

triangle BDC congruent triangle CEB by

congruence by (ASA

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