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Question

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.

Show that A>C and B>D.


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Solution

Step 1: Proving A>C.

Join AC and mark angles as 1,2,3,4, we get,

In, ABC, AB<BC (AB is the smallest side)

4<2 (Angles opposite to smaller side is smaller)…….(1)

Again, In, ADC, AD<CD (CD is the longest side)

3<1 (Angles opposite to smaller side is smaller)…….(2)

Adding equations (1) and (2), we get,

4+3<2+1 or

1+2>3+4A>C(A=1+2,C=3+4)

Hence, A>C proved.

Step 2: Proving B>D.

Join BD and mark angles as 5,6,7,8, we get,

In, ABD, AB<AD (AB is the smallest side)

6<5 (Angles opposite to smaller side is smaller)…….(3)

Again, In, CBD, BC<CD (CD is the longest side)

8<7 (Angles opposite to smaller side is smaller)…….(4)

Adding equations (3) and (4), we get,

6+8<5+7 or

5+7>6+8B>D(B=5+7,D=6+8)

Hence, B>D proved.


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