Question

D is the midpoint of the hypotenuse AC of a right angled triangle ABC which is right angled at B . Prove that BD =1/2 AC. PLEASE WRITE THE CORRECT ANSWER ON A COPY AND SEND A PHOTO. THOSE WHO GIVES NONSENCE ANSWER LIKE HLHOW ARE YOU OR ANYTHING,WILL BE REPORTED. plzz answer.

Answer 1

hello dear friend ⭐⭐⭐⭐⭐⭐
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your answer ⭐ given by Ishan
so let me tell you now ,,dear,,

Given:

Right angle triangle ΔABCΔABC where ∡ABC=90∘∡ABC=90∘

BDBD divides ACAC, i.e., AD=DCAD=DC

From DD, draw EDED and FD⊥FD⊥ to ABAB and BCBCrespectively

In DEBFDEBF, Because ∡BED = ∡BFD = ∡FBE = 90∘, ∠EDF=90∘∡BED = ∡BFD = ∡FBE = 90∘, ∠EDF=90∘. Therefore, DEBFDEBF is a rectangle. Hence, BE = DF, ED = BFBE = DF, ED = BF.

In Δs AEDΔs AED and DFCDFC,

∡EAD = ∡FDC∡EAD = ∡FDC (AB//FDAB//FD and ACAC cuts it — corresponding angles).

∡DEA = ∡CFD=90∘∡DEA = ∡CFD=90∘(Construction)

AD=DCAD=DC (Given)

So ΔAED = ΔDFCΔAED = ΔDFC. So AE=FDAE=FD. Therefore, AE=BEAE=BE.

In Δs AED,BEDΔs AED,BED,

AE=BEAE=BE

DEDE is common

included angles ∡DEA = ∡DEB=90∘∡DEA = ∡DEB=90∘(Construction)

So, ΔAED=ΔBEDΔAED=ΔBED.

Therefore, BD=AD=AC2

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