Question : If $\frac{3a+4b}{3c+4d}=\frac{3a-4b}{3c-4d}$, then:
Option 1: $ab=cd$
Option 2: $ad=bc$
Option 3: $ac=bd$
Option 4: $a=b=c\neq d$
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Correct Answer: $ad=bc$
Solution : Given: $\frac{3a+4b}{3c+4d}=\frac{3a-4b}{3c-4d}$ Simplifying by cross multiplication, $⇒9ac-12ad+12bc-16bd=9ac-12bc+12ad-16bd$ $⇒24bc=24ad$ $⇒ad=bc$ Hence, the correct answer is $ad=bc$.
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Question : ABCD is a quadrilateral in which BD and AC are diagonals. Then, which of the following is true:
Question : If ABC is an equilateral triangle and D is a point in BC such that AD is perpendicular to BC, then:
Question : If $\frac{a^{2} - bc}{a^{2}+bc}+\frac{b^{2}-ca}{b^{2}+ca}+\frac{c^{2}-ab}{c^{2}+ab}=1$, then the value of $\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ac}+\frac{c^{2}}{c^{2}+ab}$ is:
Question : ABC is a right angle triangle and $\angle ABC = 90^{\circ}$. BD is perpendicular on the side AC. What is the value of $(BD)^2$?
Question : A point D is taken from the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then:
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