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Quadrilateral $CDEF$ is a parallelogram. Its area is $36$ square units. Points $G$ and $H$ are the midpoints of sides $CD$ and $EF,$ respectively. What is the area of triangle $CDJ?$ Oct 28, 2018

#1
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$$\because \angle DHE=\angle FHJ$$ (vertical angles)

$$\because EH=HF$$ (midpoint)

$$\because \overline{DE}\parallel\overline{FJ} \Rightarrow \angle EDJ =\angle FJD$$  (alternating interior angles)

$$\therefore \triangle HDE\cong\triangle HJF$$ (AAS Congruency)

$$\therefore [HDE]=[HJF] \Rightarrow [CDHF]+[HDE]=[CDHF]+[HJF]$$

$$[CDJ]=[CDEF]$$

$$\boxed{36}$$

I hope this helped,

Gavin.

Oct 28, 2018

#1
+3

$$\because \angle DHE=\angle FHJ$$ (vertical angles)

$$\because EH=HF$$ (midpoint)

$$\because \overline{DE}\parallel\overline{FJ} \Rightarrow \angle EDJ =\angle FJD$$  (alternating interior angles)

$$\therefore \triangle HDE\cong\triangle HJF$$ (AAS Congruency)

$$\therefore [HDE]=[HJF] \Rightarrow [CDHF]+[HDE]=[CDHF]+[HJF]$$

$$[CDJ]=[CDEF]$$

$$\boxed{36}$$

I hope this helped,

Gavin.

GYanggg Oct 28, 2018