Question

can anyone answer thisthree questions at a time please.....I'll mark u as brain list​

Answer 1

Answer:

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Answer 2

21)

Take ∆ADC and ∆ABC:

AC = AC (common side)

AD = BC (opposite sides of a parallelogram are equal)

angle CAD = angle CAB (given)

Hence by SAS congruency, ∆ADC and ∆ABC are congruent.

By CPCT property, angle DCA = angle BCA.

Hence the diagonal bisects angle C.

22)

Given: x + y = w + z

Now, x = angle AOC, y = angle COB

Now, clearly x + y + w + z = 360⁰ (together they form a complete angle)

--> x + y + (x + y) = 360⁰

--> 2(x + y) = 360⁰

--> x + y = 180⁰

Hence, angle AOC + angle COB = 180⁰

Therefore, AOB is a straight line

23)

i) Consider ∆ABD and ∆BAC

AD = BC (given)

angle DAB = angle CBA (given)

AB = AB (common side)

Therefore by SAS congruency, ∆ABD and ∆BAC are congruent

ii) Now since ∆ABD is congruent to ∆BAC, therefore by CPCT, BD = AC

Hope that you will find this useful :)

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