(C) In $\Delta ABC$,$E$ is the midpoint of $\overline{AC}$.Therefore,$AC = 2EC$ $(1)$.In $\Delta BEC$,line $DK$ passes through the midpoint $D$ of $\o

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(C) In $\Delta ABC$,$E$ is the midpoint of $\overline{AC}$.Therefore,$AC = 2EC$ $(1)$.In $\Delta BEC$,line $DK$ passes through the midpoint $D$ of $\overline{BC}$ and is parallel to $\overline{BE}$.By the Converse of the Midpoint Theorem,$K$ is the midpoint of $\overline{EC}$.Therefore,$EC = 2EK$ $(2)$.Substituting $(2)$ into $(1)$,we get $AC = 2(2EK) = 4EK$.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

In $\Delta ABC$,$D$ and $E$ are the midpoints of $\overline{BC}$ and $\overline{AC}$ respectively. $\overline{AD}$ and $\overline{BE}$ intersect at $G$. Line $m$ passing through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. Then,$AC = \ldots$ (in $EK$)

A
$2$
B
$3$
C
$4$
D
$6$

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Mix Examples - Triangles

In $\Delta XYZ$,$X-P-Y$,$X-Q-Z$ and $\overline{PQ} \parallel \overline{YZ}$. If $XP = 8$,$XY = 12$ and $XQ = 12$,find $QZ$.

Medium

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In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. If $PQ = 4QR$,prove that $PD = 16RD$.

Medium

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In $\Delta ABC$,$m \angle B = 90$ and $D$ is the midpoint of $\overline{AC}$. Then,$BD = \dots$

Easy

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In $\Delta PQR$ and $\Delta XYZ$,$\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$. Which correspondence between them is a similarity?

Medium

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In $\Delta DEF, m \angle D = 90^{\circ}$. If $EF = 6$ and $DF = 4$,then $DE = \dots$

Medium

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In $\Delta ABC$,$D$ and $E$ are the midpoints of $\overline{BC}$ and $\overline{AC}$ respectively. $\overline{AD}$ and $\overline{BE}$ intersect at $G$. Line $m$ passing through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. Then,$AC = \ldots$ (in $EK$)

Similar Questions

In $\Delta XYZ$,$X-P-Y$,$X-Q-Z$ and $\overline{PQ} \parallel \overline{YZ}$. If $XP = 8$,$XY = 12$ and $XQ = 12$,find $QZ$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. If $PQ = 4QR$,prove that $PD = 16RD$.

In $\Delta ABC$,$m \angle B = 90$ and $D$ is the midpoint of $\overline{AC}$. Then,$BD = \dots$

In $\Delta PQR$ and $\Delta XYZ$,$\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$. Which correspondence between them is a similarity?

In $\Delta DEF, m \angle D = 90^{\circ}$. If $EF = 6$ and $DF = 4$,then $DE = \dots$

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