![\bold{\huge{\underline{ Solution }}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/jdi2w7914cic76zpb2xuxp7e51pz44d9g8.png)
Let consider the given triangle be ABC
According to the question,
AD is the median of the triangle ABC and CE also divides the triangle into two parts. At F both the lines are intersecting.
Here We have ,
In triangle FDC, By using Angle sum property
![\sf{ {\angle}FDC + {\angle}DCF + {\angle}CFD = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/8o208hg1tx6jaxytc0nzf36ib3vpxan8rm.png)
Subsitute the required values,
![\sf{ 90{\degree} + 20{\degree} + {\angle}CFD = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/240an359se39axxsk2sj2hrcqvnn7g4iy3.png)
![\sf{ 110{\degree} + {\angle}CFD = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/kibaak730x19ngj26keflk4gn4ujoc45bs.png)
![\sf{ {\angle}CFD = 180{\degree} - 110{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/wsypsi541z79sdq7d6c92l1v2atl6j0wwm.png)
![\sf{ {\angle}CFD = 70{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/2fsi2knz2kyofh8kxtd0wun2q65xuwo2zf.png)
Here.
![\sf{ {\angle}CFD = {\angle}EFA = 70{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/5pjuvnkcs2pu5nosn0om4rid795ys0hdik.png)
- The above angles are vertically opposite angles and vertically opposite angles are equal.
Now, Again by using Angle sum property but in triangle AEF
![\sf{ {\angle}AEF + {\angle}EFA + {\angle}FAE = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/952y5wgspwbi0ja8ujaz64tqtkpoqwgc2b.png)
Subsitute the required values,
![\sf{ x{\degree} + 70{\degree} + 25{\degree} = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/5hu2al8c40tecj1korimr1aaxq9k68t2md.png)
![\sf{ x{\degree} + 95{\degree}= 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/u1whn9o6soqjyh0tem3zi1zwxxn3omfb3i.png)
![\sf{ x{\degree} = 180{\degree} - 95{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/7s4oef4n4clngegcyx4pgs7gqpv43mepd6.png)
![\bold{ x = 85{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/75jiyhtvy9bb9czedou4mkxpq6i7j9sld5.png)
Thus , The value of x is 85°
- Here, In the triangle ABC , CE divides the triangle into two parts
So ,
![\sf{ {\angle}AEF + {\angle}FEB = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/eu3d4v91a17szb8bsifnkht9bf546gjgby.png)
![\sf{ 85{\degree} + {\angle}FEB = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/gevxskfhm61w7atw23pcqhheqvgbd6nhko.png)
![\sf{ {\angle}FEB = 180{\degree} - 85{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/uh25a0sgeqgfk5d47capkprhv94er487js.png)
![\sf{ {\angle}FEB = 95{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/nlwjs5rpwrn8ozmdfhmiidwgnl1knw5wdm.png)
Similarly ,
![\sf{ {\angle}AFE + {\angle}DFE = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/t1l4y1ylq7nxe7819m7ebnjctqgrnq2k7z.png)
![\sf{ 70{\degree} + {\angle}DFE = 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/6o49rde8c4ndu0hdvw4szflknnu1rgjc0q.png)
![\sf{ {\angle}DFE = 180{\degree} - 70{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/m94peg7k0aov1njok5twqzleban22kjg88.png)
![\sf{ {\angle}DFE = 110 {\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/bic0zfw808xy4y0ahyp65sbz3raoklqpgt.png)
- In triangle ABC, EBDF is forming quadrilateral
We know that,
- Sum of all the angles of quadrilateral is equal to 360°
That is ,
![\sf{ {\angle}FEB + {\angle}EBD + {\angle}BDF + {\angle} DFE = 360{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/yz1o266avfaey931y7djwagthbuafk9w12.png)
Subsitute the required values,
![\sf{ 95{\degree} + y{\degree} + 90{\degree} + + 110{\degree} = 360{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/v0uz7wznq7s8dozcwdy0d19mo7r1y061l6.png)
![\sf{ y{\degree} + 295{\degree}= 180{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/bf0iyw2v4vrr9xow2qmxqfccgijj7gk762.png)
![\sf{ y {\degree} = 360{\degree} - 295{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/e9yzljg50c8b2rig2up2xahynmt4j6u7vz.png)
![\bold{ y = 65{\degree}}](https://img.qammunity.org/2023/formulas/mathematics/college/parnr8klh1qhjynqbegz6ihkunrwo1ybfr.png)
Hence, The value of x and y is 70° and 65° .