Find Measure of ∠BAD in ABCD Rhombus

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Discussion Overview

The discussion revolves around determining the measure of angle ∠BAD in a rhombus ABCD, with specific points H and K defined on sides BC and CD, respectively. Participants explore geometric relationships and angle measures within the context of this rhombus configuration.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant defines a rhombus and sets up the problem involving points H and K, suggesting the use of isosceles triangles to find angle measures.
  • Another participant proposes denoting angle B as x and expresses a method to relate other angles to x, aiming to derive an equation involving x.
  • A later reply claims to have found angle A to be 100 degrees, although the method leading to this conclusion is not detailed.
  • Another participant provides a calculation that leads to the conclusion that angle B is 80 degrees, subsequently stating that angle BAD is 100 degrees.

Areas of Agreement / Disagreement

There is no consensus on the measure of angle BAD, as different participants provide varying calculations and results. The discussion remains unresolved regarding the correctness of the angle measures proposed.

Contextual Notes

Participants have not fully detailed their assumptions or the steps leading to their conclusions, leaving some mathematical relationships and dependencies on definitions unclear.

Ilikebugs
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A parallelogram is a quadrilateral with opposite sides parallel. A rhombus is a parallelogram with all four sides having equal length.

ABCD is a rhombus. H is on BC, between B and C, and K is on CD, between
C and D, such that AB = AH = HK = KA.
Determine the measure of ∠BAD.I've tried making isosceles triangles with little success
 
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\begin{tikzpicture}
\renewcommand\ss{2}
\coordinate[label=above:$A$] (A) at (0,0);
\path (A) ++(220:\ss) coordinate[label=left:$D$] (D);
\path (A) ++(-40:\ss) coordinate[label=right:$B$] (B);
\path (D) ++(-40:\ss) coordinate[label=below:$C$] (C);
\path (A) ++(-60:\ss) coordinate[label=below right:$H$] (H);
\path (A) ++(240:\ss) coordinate[label=below left:$K$] (K);
\draw (A) -- (B) -- (C) -- (D) -- cycle;
\draw (A) -- (H) -- (K) -- cycle;
\foreach \p in {A,B,D,C,H,K} \fill (\p) circle (1.5pt);
\end{tikzpicture}​
I suggest denoting $\angle B=\angle D$ by $x$ and trying to express other angles through it, eventually arriving at some equation in $x$. Note that the sides of the rhombus and the sides of the triangle are all equal.
 
I got A=100 degrees?
 
Last edited:
$x=\angle{B},\quad2x+4(180-2x)+120=360\implies x=80^{\circ}\implies\angle{BAD}=100^{\circ}$
 

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