Question

In triangle abc ,AD is perpendicular to BC . Ad is called------- of the triangle

Answer 1

Step-by-step explanation:

It could be bisecter.

hope this answer helps you dear!

Answer 2

Answer:

To prove : BAC = 90°

To prove : BAC = 90°in right triangles ∆ADB and ∆ADC

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be applied

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we have

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = AD² + BD² ———-(1)

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = AD² + BD² ———-(1)AC²= AD²+ DC² ———(2)

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = AD² + BD² ———-(1)AC²= AD²+ DC² ———(2)AB² + AC² = 2AD² + BD²+ DC²

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = AD² + BD² ———-(1)AC²= AD²+ DC² ———(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = AD² + BD² ———-(1)AC²= AD²+ DC² ———(2)AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]= (BD + CD )² = BC²

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = 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

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = 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

To prove : BAC = 90°in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be appliedThen we haveAB² = 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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