Question

In the figure AB || DE and BD || EF. Prove that DC²= CF x AC​

Answer 1

Hey mate your answer is in the given attachment

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Answer 2

By Thales' Theorem,

★ since AB || DE, in ∆ABC,

$\displaystyle\longrightarrow\sf {\dfrac {AC}{DC}=\dfrac {BC}{CE}\quad\quad\dots (1)}$

★ since BD || EF, in ∆BCD,

$\displaystyle\longrightarrow\sf {\dfrac {DC}{CF}=\dfrac {BC}{CE}\quad\quad\dots (2)}$

From (1) and (2),

$\displaystyle\longrightarrow\sf {\dfrac {AC}{DC}=\dfrac {DC}{CF}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {(DC)^2=CF\cdot AC}}}$

Hence Proved!

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Thales' Theorem

It states that, a line drawn parallel to one side of a triangle in it cuts the other two sides in the same ratio as that kept by this line with the parallel side of the triangle.

Proof:- Let us consider ∆ABC as the reference in which it is given that AB || DE.

Consider ∆ABC and ∆CDE.

$\displaystyle\longrightarrow\sf {\angle ACB=\angle DCE\quad (common)}$

$\displaystyle\longrightarrow\sf {\angle BAC=\angle CDE\quad (corresponding\ angles)}$

Therefore, $\displaystyle\sf {\triangle ABC=\triangle CDE}$

And hence,

$\displaystyle\longrightarrow\sf {\dfrac {CD}{AC}=\dfrac {CE}{BC}=\dfrac {DE}{AB}}$

QED