In - ABC- AD is perpendicular to BC- Then

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Magnetic Field Due to Current in a Solenoid

In Δ ABC, A...

Question

In $Δ A B C, A D$ is perpendicular to $B C$ . Then

A D^{2} - D C^{2} = A C^{2} + B D^{2}

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A B^{2} - A C^{2} = B C (B D - C D)

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A B^{2} - A D^{2} = B C (B C - C D)

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B C^{2} - C D^{2} = B D (A C + B D)

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Similar questions

In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

(a) $B C^{2} + A D^{2} + 2 B C . A D$

(b) $A B^{2} + C D^{2} + 2 A B . C D$

(d) $B C^{2} + A D^{2} + 2 A B . C D$

△ A B C

{A B}^{2} + {C D}^{2} = {A C}^{2} + {B D}^{2}

△ A B C

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A C^{2} = A D^{2} + B C \times D M + \frac{B C^{2}}{4}

A B^{2} = A D^{2} - B C \times D M + \frac{B C^{2}}{4}

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