Class 10 Exam > Class 10 Questions > In right triangle ABC, right angled at A,A pe...
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In right triangle ABC, right angled at A,
A perpendicular is dropped from A to BC, meeting BC at D. Then which of the following is true?
- a)ΔADC ~ ΔABD
- b)ΔDCA ~ ΔDABD
- c)ΔDAC ~ ΔDABD
- d)ΔDAC ~ ΔDABA
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
In right triangle ABC, right angled at A,A perpendicular is dropped fr...
Explanation:
- Let's draw the diagram first.
- From the diagram, we can see that triangle ABD and triangle ACD are both right triangles.
- Therefore, we can use the Pythagorean theorem to find their sides.
- Let's assume that AB = b, AC = c, and BC = a.
- Using Pythagorean theorem, we get:
- AB² + BD² = AD² (for triangle ABD)
- AC² + CD² = AD² (for triangle ACD)
- BC² = AB² + AC² (by Pythagoras theorem)
- Now, we can simplify the above equations to get:
- BD² = AD² - AB² = (AC² + CD²) - AB²
- CD² = AD² - AC² = (AB² + BD²) - AC²
- Substituting the value of BD² and CD² in the above equations, we get:
- AB² + (AC² + CD² - AB²) = AD²
- AC² + (AB² + BD² - AC²) = AD²
- After simplifying, we get:
- AC² = AD² - AB²
- AB² = AD² - AC²
- Therefore, we can say that triangle DAC is similar to triangle DAB by the Angle-Angle-Similarity criterion.
- Thus, option D is the correct answer.
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Community Answer
In right triangle ABC, right angled at A,A perpendicular is dropped fr...
∆DAC~∆ABC)......1eq angleD=angA]right angle angC=angC]common similarly ∆DAB~∆ACB).....2eq therefore ∆DAC~∆DAB
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