Question

AB = AC + BC
(By using Pythagoras Theorem]
AC = BC
...(Given)
AB=AC+ AC
AB²=2AC
Hence, proved.
Q. 11. Prove that the sum of the squares of the sides of a rhombus is equal to the sum
the squares of its diagonals.
(P.S.E.B. 2019 Set A, B.
Salution. Given: A rhombus ABCD whose diagonals AC and BD intersect each other at​

Answer 1

Answer:

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Answer 2

Answer:

Given:- A right triangle ABC right angled at B.

To prove:- AC2  =AB2  +BC2

 Construction:- Draw BD⊥AC

Proof:-

In △ABC and △ABD

∠ABC=∠ADB(Each 90°)

∠A=∠A(Common)

∴△ABC∼△ABD(By AA)

AB /AC  =AD /AC  (∵Sides of similar triangles are proportional)

⇒AB2  =AD⋅AC.....(1)

Similarly, in △ABC and △BCD

∠ABC=∠BDC(Each 90°)

∠C=∠C(Common)

∴△ABC∼△BCD(By AA)

∴ DC / BC =  BC /AC

⇒BC2  =DC⋅AC.....(2)

Adding equation (1)&(2), we have

AB2  +BC2  =AD⋅AC+DC.AC

⇒AB2  +BC2  =AC(AD+DC)

⇒AB2+BC2 =AC2

Hence proved.

Step-by-step explanation: