Question

help:
figure and hint is given above

A quadrilateral ABCD is such that the diagonal AC bisects each of the angles BAD and BCD . Prove that AC bisects BD at right angles. ​

Answer 1

Step-by-step explanation:

Okay here it is :-

In triangle ABC and ADC,

Angle CAB = Angle CAD ( Since AC bisect Angle BAD)

Angle BCA = Angle DCA ( Since AC bisect Angle BCD)

AC = AC ( Common)

So, Triangle ABC is congruent to triangle ADC ( By A.A.S)

Therefore, BC = DC ( By C.P.C.T)

Now, In triangle BEC and DEC,

Angle ECB = Angle ECD = x ( Since AC bisect Angle BCD)

BC = DC ( Since triangle ABC is congruent to triangle ADC)

EC = EC ( Common )

So, Triangle BEC is congruent to DEC.( By S.A.S)

Therefore, Angle BEC = Angle DEC ( By C.P.C.T)

Also, Angle BEC + Angle DEC = 180° ( linear pair)

=> 2( Angle BEC) = 180° ( Since Angle BEC = Angle DEC)

So, Angle BEC = 180°/2

= 90°

Also, Angle DEC = Angle BEC = 90°

Since, Angle DEC = Angle BEC = 90°

Therefore, AC bisects BD at right angles. Proved✓

Hope it helps

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Answer 2

Answer:

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