Question

In a quad. ABCD ANGLE A + ANGLE D =90 DEGREE. THEN PROVE THAT AC^2 + BD^2=BC^2 + AD^2.(Answer it up please)

Answer 1

${\underline{ \mathtt{\red{A} \green{n}\mathtt\blue{s} \purple{w} \mathtt \orange{e}\pink{r}}}}\:$

Given  

→ To prove $AC^2+BD^2=AD^2+BC^2$

⊕ PRODUCE AB AND CD TO MEET AT E

⊕ ALSO JOIN AC AND BD,

⊕ PROOF,

⊕ IN TRINGLE AED WE HAVE  

→ angle A+angle D = 90 degree

→ angle E = $180-90$  degree (angle sum property of triangle)

⊕ by pythagoras theorem  

→ $AD^2=AE^2+DE^2$

⊕ IN TRIANGLE BEC BY PHYTHAGORAS THEOREM

→ $BC^2=BE^2+EF^2$

⊕ ON ADDING BOTH EQUATIONS WE GET

→ $AD^2+BC^2=AE^2+BE^2+DE^2+EF^2$                           (1)

⊕ IN TRIANGLE AEC BY PHYTHAGORAS THEOREM

→ $AC^2=AE^2+CE^2$

⊕ AND IN TRIANGLE BED BY PYTHAGORAS THEOREM

→ $BD^2=BE^2+DE^2$

⊕ ON ADDING BOTH EQUATIONS WE GET

→ $AC^2+BD^2=AE^2+BE^2+CE^2+DE^2$                                                  (2)

⊕ FROM EQUATIONS 1 AND 2 WE GET  

→ $AC^2+BD^2=AD^2+BC^2$