Question

In a triangle \(ABC\), \(AB=3\), \(BC=4\) and \(CA=5\). Point \(D\) is the midpoint of \(AB\), point \(E\) is on segment \(AC\) and point \(F\) is on segment \(BC\). If \(AE=1.5\) and \(BF=0.5\), then \(\angle DEF =\)

• A. 30 degrees
• B. 45 degrees
• C. 60 degrees
• D. 75 degrees

Hint:

Spot the two isosceles triangles formed at A and at C using the given lengths, before trying to chase angles at E.

Answer 1

The Correct Option is B

Step 1: Spot the right angle.
Since \(AB=3\), \(BC=4\), \(CA=5\), and \(3^2+4^2=9+16=25=5^2\), triangle \(ABC\) is right-angled at \(B\) (the side opposite the largest side \(CA\) is opposite the right angle), so \(\angle ABC=90^{\circ}\).

Step 2: Work out the two extra lengths on the triangle.
\(D\) is the midpoint of \(AB\), so \(AD=DB=1.5\). Since \(AE=1.5\), we get \(AD=AE=1.5\), so triangle \(ADE\) is isosceles with the equal sides at \(A\). Also, \(CE=CA-AE=5-1.5=3.5\) and \(CF=BC-BF=4-0.5=3.5\), so \(CE=CF=3.5\), meaning triangle \(CEF\) is isosceles too, with the equal sides at \(C\).

Step 3: Name the base angles of these isosceles triangles.
In isosceles triangle \(ADE\), the base angles are equal: let \(\angle AED=\angle ADE=x\). In isosceles triangle \(CEF\), the base angles are equal: let \(\angle CEF=\angle CFE=z\). Let \(\angle DEF=y\), the angle we want.

Step 4: Use the straight line at E.
Points \(A\), \(E\), \(C\) are collinear (E lies on segment AC), so the three angles at \(E\) on that side add up to \(180^{\circ}\): \[ x+y+z=180^{\circ} \qquad (ii) \]

Step 5: Use the angle sum of quadrilateral BDEF.
Consider quadrilateral \(BDEF\); its interior angles add up to \(360^{\circ}\). At \(D\), the interior angle of the quadrilateral is \(180^{\circ}-x\); at \(F\), it is \(180^{\circ}-z\); at \(B\), it is \(\angle DBF=\angle ABC=90^{\circ}\); at \(E\), it is \(\angle DEF=y\). So \[ y + (180^{\circ}-z) + 90^{\circ} + (180^{\circ}-x) = 360^{\circ} \] \[ 450^{\circ}+y-(x+z) = 360^{\circ} \]

Step 6: Substitute from equation (ii).
From (ii), \(x+z=180^{\circ}-y\). Substituting: \[ 450^{\circ}+y-(180^{\circ}-y)=360^{\circ} \] \[ 270^{\circ}+2y=360^{\circ} \] \[ 2y=90^{\circ} \] \[ y=45^{\circ} \]

Final Answer:
\(\angle DEF=45^{\circ}\). \[ \boxed{45^{\circ}} \]