# Let hat(ABC) be any triangle, stretch bar(AC) to D such that bar(CD)≅bar(CB); stretch also bar(CB) into E such that bar(CE)≅bar(CA). Segments bar(DE) and bar(AB) meet at F. Show that hat(DFB is isosceles?

## Given any triangle $\hat{A B C}$, stretch $\overline{A C}$ to D such that bar(CD)≅bar(CB); stretch also $\overline{C B}$ into E such that bar(CE)≅bar(CA).Segment $\overline{D E} \mathmr{and} \overline{A B}$ meet at F. Show that hat(DFB is isoscheles? Jun 19, 2016

As follows

#### Explanation:

Ref:Given Figure

$\text{In } \Delta C B D , \overline{C D} \cong \overline{C B} \implies \angle C B D = \angle C D B$

$\text{Again in } \Delta A B C \mathmr{and} \Delta D E C$

$\overline{C E} \cong \overline{A C} \to \text{by construction}$

$\overline{C D} \cong \overline{C B} \to \text{by construction}$

$\text{And "/_DCE=" vertically opposite } \angle B C A$

$\text{Hence } \Delta A B C \cong \Delta D C E$
$\implies \angle E D C = \angle A B C$

$\text{Now in } \Delta B D F , \angle F B D = \angle A B C + \angle C B D = \angle E D C + \angle C D B = \angle E D B = \angle F D B$

$\text{So "bar(FB)~=bar(FD)=>DeltaFBD " is isosceles}$
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