Points $A, B$ are on the same side of a line $l$ . $A D ⊥ l$ , and $B E ⊥ l$ meet $l$ in $D$ and $E$ respectively. $C$ is the mid-point of $A B$ (see in the figure). Prove that $C D = C E$ .

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Solution

line $A D ⊥$ line l (given)
line $B E ⊥$ line l (given)

Draw line $C M ⊥$ line l -----(1)
lines $A D | | C M | | B E$

AC=CB (given (since C is mid point AB)

Since CM, is traversal to lines AD and BE
Hence we can say that DM=ME
In $Δ D C M$ and $Δ M C E$

$D M = M E$ from
CM is te common line (common side)$

$D M C = ∠ E M C = 90^{o}$ from

$Δ D C M ≅ Δ M C E$

$C D = C E$

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