Question

plz tell me how to solve these two question .........

Answer 1

HERE IS YOUR PERFECT ANSWER:
QUESTION 1. SOLUTION
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GIVEN: AC = AE,
AB = AD,
Angle BAD = Angle EAC.
TO PROVE: BC = DE
PROOF:Angle BAD = Angle EAC (Given)
Therefore, Angle BAD + Angle DAC =
Angle DAC + Angle CAE.
→ Angle BAC = Angle DAE.
In ∆BAC and ∆DAE,
AB = AD (Given)
Angle BAC = Angle DAE ( Just Proved )
AC =AE (Given)
Therefore, ∆BAC is congruent to ∆DAE
( By S.A.S )
So, BC = DE ( By C.P.C.T )
Hence Proved.

QUESTION 2. SOLUTION.
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GIVEN: Angle ACB = 90° ;
AM = BM ;
DM = CM.
TO PROVE: 1} ∆AMC is congruent to ∆BMD.
PROOF: In ∆AMC and ∆BMD,
AM = BM (Given)
CM = DM (Given)
Angle AMC = Angle BMD (V.O.A)
Therefore,∆AMC is congruent to ∆BMD
(By S.A.S)
2} Angle DAC = 90°.
PROOF: Angle BDM = Angle AMC ( By C.P.C.T )
Therefore, AC and BD are parallel
because these angles are alternate
interior angles.
Angle DBC + Angle ACB =180°
(Both are right angles)
Angle DBC = 180° - 90°
Angle DBC = 90°.
3} ∆DBC is congruent to ∆ACB.
PROOF: In ∆DBC and ∆ACB,
DB = AC (From part {1} by C.P.C.T )
Angle DBC = Angle ACB = 90°.
BC = BC (Common)
Therefore, ∆DBC is congruent to ∆ACB
(By S.A.S)
So, DC = AB ( By C.P.C.T )
4} CM = 1/2 AB
PROOF: 1/2DC = 1/2AB
So, CM = 1/2AB

Hence Proved.

HOPE THIS WILL HELP YOU.
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