in quadrilateral abcd, angle a+angle b =90°.prove that ab²+cd²=ac²+bd².
Answers
Answer:
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Answer:
AS shown in the attached diagram, ABCD is a quadrilateral, with ∠A + ∠D = 90°.
We are to prove that AC² + BD² = AD² + BC².
The sides AB and DC are produced to meet at "P".
Now, in ΔAPD, we have
\begin{gathered}\angle A+\angle P+\angle D=180^\circ\\\\\Rightarrow \angle P=180^\circ-90^\circ\\\\\Rightarrow \angle P=90^\circ.\end{gathered}
∠A+∠P+∠D=180
∘
⇒∠P=180
∘
−90
∘
⇒∠P=90
∘
.
Using Pythagoras Theorem in ΔAPD, ΔBPC, ΔAPC and ΔBPD, we have
\begin{gathered}AD^2=AP^2+PD^2,\\\\BC^2=BP^2+PC^2,\\\\AC^2=AP^2+PC^2,\\\\BD^2=DP^2+BP^2.\end{gathered}
AD
2
=AP
2
+PD
2
,
BC
2
=BP
2
+PC
2
,
AC
2
=AP
2
+PC
2
,
BD
2
=DP
2
+BP
2
.
Now,
\begin{gathered}R.H.S.\\\\=AD^2+BC^2\\\\=AP^2+PD^2+BP^2+PC^2\\\\=(AP^2+PC^2)+(PD^2+BP^2)\\\\=AC^2+BD^2\\\\=L.H.S.\end{gathered}
R.H.S.
=AD
2
+BC
2
=AP
2
+PD
2
+BP
2
+PC
2
=(AP
2
+PC
2
)+(PD
2
+BP
2
)
=AC
2
+BD
2
=L.H.S.
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