Question

In ∆ ABC , AD perpendicular BC and AD² = BD • CD.prove that angle BAC = 90°.

Answer 1

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

To prove : BAC = 90°

Proof : in right triangles ∆ADB and ∆ADC
So, 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
Thus in triangle ABC we have , AB² + AC²= BC²

hence triangle ABC is a right triangle right angled at A

∠ BAC = 90°

Hope it will help you‼️

Answer 2

Given : A ∆ABC in which AD perpendicular BC and AD²= BD × CD.

To Prove :- /_ BAC = 90°


Proof :- AD² = BD × CD => BD / AD = AD/CD.



Now , in ∆ DBA and ∆ DAC , we have :

/_ BDA = /_ ADC = 90° and BD / AD = AD / CD


Therefore,


∆ DBA ~ ∆ DAC [ SAS - Similarity ]


Therefore,


/_ B = /_ 2 and /_ 1 = /_ C



/_ 1 + /_2 = /_ B + /_ C => /_ A = /_ B + /_ C





=> 2/_ A = /_ A + /_ B + /_ C = 180°


=> /_ A = 90°.