if in a right angle triangle ABC , right angled at A, AD perpendicular to BC, then prove that AB²+CD²=BD²+AC²
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Step-by-step explanation:
In a Triangle ABC with AD perpendicular to BC, two right angled Triangles are formed Viz. Triangle ADB and Triangle ADC.
Since these are Right Angled triangles, we have
AB^2 = BD^2 + AD^2 —— 1 and
AC^2 = CD^2 + AD^2 —— 2
From 1, we get
AD^2 = AB^2 - BD^2 —— 3
and from 2, we get
AD^2 = AC^2 - CD^2 —— 4
Since both 3 and 4 are about AD^2, we get
AB^2 - BD^2 = AC^2 - CD^2
On rearranging BD^2 and CD^2, we get
AB^2 + CD^2 = AC^2 + BD^2
Hence proved!
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