Question

triangle ABC is a right angled at a and a d perpendicular BC divides BC such that ad square equal to BD into CD if ad is equal to 4 cm and BD equal to 2 cm find BC​

Answer 1

In triangle ABC, AD perpendicular to BC and AD2 =BD * CD.prove that triangle ABC is a right triangle

Answer 2

" ✌️ Answer ✌️ "

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DC

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADC

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)AB² + AC² = 2AD² + BD²+ DC²

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]= (BD + CD )² = BC²

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]= (BD + CD )² = BC²Thus in triangle ABC we have , AB² + AC²= BC²

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]= (BD + CD )² = BC²Thus in triangle ABC we have , AB² + AC²= BC²hence triangle ABC is a right triangle right angled at A

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DCTo prove : BAC = 90°Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply ,Then we have ,AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]= (BD + CD )² = BC²Thus in triangle ABC we have , AB² + AC²= BC²hence triangle ABC is a right triangle right angled at A∠ BAC = 90°

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