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# question

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Let ABCD be a parallelogram. We have that M is the midpoint of AB and N is the midpoint of BC. The segments DM and DN intersect AC at P and Q, respectively. If AC = 15, what is QA?

Mar 26, 2018

#1
+24983
+3

Let ABCD be a parallelogram. We have that M is the midpoint of AB and N is the midpoint of BC.

The segments DM and DN intersect AC at P and Q, respectively.

If AC = 15, what is QA?

$$\text{Let NC = \dfrac{AD}{2}  }$$

intercept theorem:

$$\begin{array}{|rcll|} \hline \dfrac {QC}{NC} &=& \dfrac{QA}{AD} \quad & | \quad QC = AC - QA \\\\ \dfrac {AC - QA }{NC} &=& \dfrac{QA}{AD} \\\\ (AC - QA )\cdot AD &=& NC \cdot QA \quad & | \quad NC = \dfrac{AD}{2} \\\\ (AC - QA )\cdot AD &=& \dfrac{AD}{2} \cdot QA \quad & | \quad : AD \\\\ AC - QA &=& \dfrac{ QA}{2} \\\\ AC &=& QA + \dfrac{ QA}{2} \\\\ AC &=& \dfrac{3}{2}QA \\\\ \dfrac{3}{2}QA &=& AC \\\\ QA &=& \dfrac{2}{3}AC \quad & | \quad AC = 15 \\\\ QA &=& \dfrac{2}{3}\cdot 15 \\\\ QA &=& 2\cdot 5 \\\\ \mathbf{QA} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}$$

Mar 26, 2018
edited by heureka  Mar 26, 2018

#1
+24983
+3

Let ABCD be a parallelogram. We have that M is the midpoint of AB and N is the midpoint of BC.

The segments DM and DN intersect AC at P and Q, respectively.

If AC = 15, what is QA?

$$\text{Let NC = \dfrac{AD}{2}  }$$

intercept theorem:

$$\begin{array}{|rcll|} \hline \dfrac {QC}{NC} &=& \dfrac{QA}{AD} \quad & | \quad QC = AC - QA \\\\ \dfrac {AC - QA }{NC} &=& \dfrac{QA}{AD} \\\\ (AC - QA )\cdot AD &=& NC \cdot QA \quad & | \quad NC = \dfrac{AD}{2} \\\\ (AC - QA )\cdot AD &=& \dfrac{AD}{2} \cdot QA \quad & | \quad : AD \\\\ AC - QA &=& \dfrac{ QA}{2} \\\\ AC &=& QA + \dfrac{ QA}{2} \\\\ AC &=& \dfrac{3}{2}QA \\\\ \dfrac{3}{2}QA &=& AC \\\\ QA &=& \dfrac{2}{3}AC \quad & | \quad AC = 15 \\\\ QA &=& \dfrac{2}{3}\cdot 15 \\\\ QA &=& 2\cdot 5 \\\\ \mathbf{QA} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}$$

heureka Mar 26, 2018
edited by heureka  Mar 26, 2018
#2
+111329
0

Nice, heureka!!!....one question....what is the  intercept theorem???.....I've seen you refer to this several times in the past.....

CPhill  Mar 26, 2018