Question

plz solve this question....

Answer 1

hey mate
here's the solution

Answer 2

Let ∠DMB=∠1

∠AMC=∠2

∠BAC=∠4

∠ABD=∠3

Given: ∠C=90°

AM=BM

DM=CM

To prove: (i) ΔAMC≅ΔBMD

(ii) ∠DBC=90°

(iii) ΔDBC≅ΔACB

(iv) CM=1/2AB

Prove: (i) In ΔAMC and ΔBMD

AM=BM (given)

MC=MD(given)

∠1=∠2(V.O.A's)

∴ΔAMC≅ΔBMD (by SAS)

and so AC=DB and ∠4=∠3 (by cpct)

(ii) ∠4=∠3(proved)[AIA's]

∴DB║AC

AC║BD and BC is the transversal

∠C+∠B=180°

∴ ∠B=180°-90°=90°

(iii) In ΔDBC and ΔACB

DB=AC(proved)

∠B=∠C(each 90°)

BC=CB(common)

∴ΔDBC≅ΔACB( by SAS)

and so DC=AB (by cpct)

(iv) DC=AB (proved)

1/2DC=1/2AB

CM=1/2AB       Proved